Intereting Posts

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Any good decomposition theorems for total orders?
arbitrary large finite sums of an uncountable set.
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Show that $|z_1 + z_2|^2 < (1+C)|z_1|^2 + \left(1 + \frac{1}{C}\right) |z_2|^2$
If $\gcd(a,b)=1$ then, $\gcd(a^2,b^2)=1$
Dual residual for linearized ADMM
Probability Calculations on Highway

The recursion formula

$$\sum_{k=0}^{n-1}{n\choose k}B_k=0$$

which is equation (34) from the MathWorld page is the most basic bernoulli number recursion.

- Original author of an exponential generating function for the Bernoulli numbers?
- Generalizing the sum of consecutive cubes $\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$ to other odd powers
- Sum Involving Bernoulli Numbers : $\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$
- Explicit formula for Bernoulli numbers by using only the recurrence relation
- Is $\frac{\zeta (m+n)}{\zeta (m)\zeta (n)}$ a rational number for $m,n\ge 2\in\mathbb N$?
- Prove the von Staudt-Clausen congruence of the Bernoulli numbers

Another recursion is given by the following:

Consider

$$\mathbf{A} = \begin{bmatrix}

-\frac{1}{2} & -\frac{1}{6} & -\frac{1}{12} & -\frac{1}{20} & -\frac{1}{30} & -\frac{1}{42}\\

\frac{1}{1} & 0 & 0 & 0 & 0 & 0\\

0 & \frac{1}{2} & 0 & 0 & 0 & 0\\

0 & 0 & \frac{1}{3} & 0 & 0 & 0\\

0 & 0 & 0 & \frac{1}{4} & 0 & 0\\

0 & 0 & 0 & 0 & \frac{1}{5} & 0 \end{bmatrix}$$

This matrix was constructed by putting $-1/(1\cdot2), -1/(2\cdot3), -1/(3\cdot4), …$ on the first row and $1/1, 1/2, 1/3, …$ on the subdiagonal.

Now with $k!A^ke$ with $e:=[1, 0, 0, 0, 0, 0]^T$ and $k < 6$ we get the coefficients of the kth Bernoulli polynomial. With the constant term in the ‘first’ (leftmost) position. For example

$$3!A^3e = [0, 1/2, -3/2, 1, 0, 0]^T$$

and

$$4!A^4e = [-1/30, 0, 1, -2, 1, 0]^T$$

Together with this equation from wikipedia

$$B_n(x) = \sum_{k=0}^n{n\choose k}B_kx^{n-k}$$

we are given another bernoulli number recursion.

If these two recursions are related – in what way exactly?

- How to find sums like $\sum_{k=0}^{39} \binom{200}{5k}$
- How can I prove the formula for calculating successive entries in a given row of Pascal's triangle?
- Prove that $\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}$.
- Evaluate $ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots$
- Sum with binomial coefficients: $\sum_{k=0}^{n}{2n\choose 2k}$
- Prove that ${2^n-1\choose k}$ and ${2^n-k\choose k}$ ar always odd.
- Evaluate $\sum_{k = 0}^{n} {n\choose k} k^m$
- Finding a combinatorial argument for an interesting identity: $\sum_k \binom nk \binom{m+k}n = \sum_i \binom ni \binom mi 2^i$
- Give the combinatorial proof of the identity $\sum\limits_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$
- Counting elements in cartesian power with plurality + pattern constraints

Perhaps the thoughts which I followed recently give a satisfactory answer. I’ve looked at the “ZETA”-matrix and fiddled out the connection to the integral-representation in the Euler-MacLaurin-Formula. The article is not yet ready, needs some brushing and completing references and such. But as it might be helpful here: here is the link the integral in Euler-MacLaurin in connection with the Pascal-matrix

[update]: added the re-translation into bernoulli-numbers in the final formulae in the text

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